Problem: Factor completely. $96-6x^2=$
Answer: First, we take a common factor of $6$. $96-6x^2=6(16-x^2)$ Now, let's factor $16-x^2$. Both $16$ and $x^2$ are perfect squares, since $16=({4})^2$ and $x^2=({x})^2$. $16-x^2 = ({4})^2-({x})^2$ So we can use the difference of squares pattern to factor. ${a}^2 - {b}^2 =({a}+{b})({a}-{b})$ In this case, ${a}={4}$ and ${b}={x}$ : $({4})^2 - ({x})^2 =({4}+{x})({4}-{x})$ $\begin{aligned} 96-6x^2&=6(16-x^2) \\\\ &=6(4+x)(4-x) \end{aligned}$ In conclusion, the complete factorization is $6(4+x)(4-x)$ Remember that you can always check your factorization by expanding it.